Prove that $b^2-4ac$ can not be a perfect square Given $a$,$b$,$c$ are odd integers Prove that $b^2-4ac$ can not be a perfect square.
My try:Let $a=2k_1+1,b=2n+1,c=2k_2+1;n,k_1,k_2 \in I$
$b^2-4ac=(2n+1)^2-4(2k_1+1)(2k_2+1)$
$\implies b^2-4ac=4n^2+4n+1-16k_1k_2-8k_2-8k_1-4  $
 A: Assume $b^2-4ac=d^2$.
Then $d$ is odd and $(b-d)(b+d)=4ac$.
Consequently, there exists odd $u,v$ such that $b-d=2u$ and $b+d=2v$.
This leads to the contradiction $b=u+v$ since $u+v$ is even.
A: Suppose $b^2 - 4ac$ were a perfect square. Then the quadratic $ax^2 + bx + c$ has a rational root by the quadratic formula. By the rational roots theorem, the root can be written as $\frac{p}{q}$ with $p$ a factor of $c$, $q$ a factor of $a$. In particular, $p$ and $q$ are both odd. Plugging in, we have
$a (\frac{p}{q})^2 + b \frac{p}{q} + c = 0$
$a p^2 + b pq + c q^2 = 0$
All three of $ap^2$, $bpq$, and $cq^2$ are odd. And the sum of three odd numbers is odd. Thus, zero is odd. Contradiction.
Thus, $b^2 - 4ac$ can't be a perfect square.
A: You had $b^2-4ac=4n(n+1)-16k_1k_2-8k_2-8k_1-3$.
This is clearly odd, so if it were a square, it would be the square of an odd number, say $d=2j+1$,
and we'd have $4n(n+1)-16k_1k_2-8k_2-8k_1-3=4j(j+1)+1$.
But that can't be, because $n(n+1)$ and $j(j+1)$ are even, so $8$ divides $4n(n+1)$ and $4j(j+1)$,
so this would imply $8$ divides $1+3=4$, a contradiction.
If you know modular arithmetic, this argument is easier.
